Question

For any posisitive integer n,prove that n cube -n is divisible by 6.​

Answer 1

first of all

n³-n can be taken as n(n²-1)

n(n-1)(n+1)

now put n as (6q),(6q+1),(6q+2),(6q+3),(6q+4),(6q+5)

as every positive integer is of the form 3q+r

so case 1

n=3q

n³-n=n(n-1)(n+1)=3q(3q-1)(3q+1)

=3(q)(3q+1)(3q+2)

so it is divisible by 3

now u can take case 2 as n=3q+1

and all the 6 cases

you will find that n³-n is divisible in all cases

so

$\boxed{n {}^{3} - n \: is \: always \: divisble \: by \: 3}$

hope it helps

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